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1 Reversible visible watermarking and lossless recovery of original images Source: IEEE transactions on circuits and systems for video technology, vol. 16, no. 11, November 2006 Authors: Yongjian Hu and Byeungwoo Jeon Speaker: Chia-Chun Wu ( 吳佳駿 ) Date: 2007/05/02
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2 Outline Introduction Reversible visible watermarking Proposed method Embedding Lossless Recovery Experimental results Conclusions Our proposed method
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3 Introduction (1/2) Generally, a visible watermark is translucently laid on the host image and designed to be irreversible so as to resist unintentional modifications or malicious attacks.
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4 Introduction (2/2) However, in some potential applications, a visible watermark is first used as a tag or ownership identifier and then needs to be removable. Example: 1. patient’s images 2. remote sensing 3. military imagery
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5 Reversible visible watermarking Embedding algorithm Host image Marked-image Extraction algorithm Host imageMarked-image recover SKKU User key (80 bits)
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6 The Proposed method − Embedding (1/5) User key (80 bits) = Watermark size (8 bits+8 bits) + Origin position of R (16 bits+16 bits) + D c size (16 bits) + key bit plane level (3 bits) … H=S c ∪ D c W =Binary watermark S =Pixel sequence composed of one-bit pixels on I k Fig. 1. Framework of visible watermark embedding and data hiding.
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7 The Proposed method − Embedding (2/5) 148150146 147148147146149 148149 150149148 149146 150 151152153150148151147 149 146148147149146 145143140144142144145140148 144142143144146148149143150 146145148 147146145148149 148149 150149148 149146 148 147150151153147 146 R I-R Host Image I 011 001 010 Watermark W
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8 The Proposed method − Embedding (3/5) 149146148 144142144 146148 R 011 001 010 W 100101011001001010010100 100100001000111010010000 1001001010010100 21146148 1614144 1614620 R’ 000101011001001010010100 000100000000111010010000 000100001001001000010100 111 111 111 D D c = 010 Hiding Compression
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9 The Proposed method − Embedding (4/5) To find S which satisfy |D c |=|S|-|S c | D c = 010, |D c | = 3 000010101 011010010 001010011 111110 110R’100 001110 010010101 011010010 001011110 LSB of I-R 0 S=0,|S|=1, |S c |=1 X 0 S=00, |S|=2, |S c |=2 X 0 S=000, |S|=3, |S c |=2 X S=000010101011 S c =001001011 |S|=12, |S c |=9 O 000010101 011 Payload: H=S c ∪ D c H=001001011 010 001001011 010
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10 The Proposed method − Embedding (5/5) 148150147146 149146147149 148149148150149148 149146 150 151152153150148151147 149 21146148147149146 1451431401614144145140148 1441421431614620149143150 146145148 147146145148149 148149 150149148 149146 148 147150151153147 146 RmRm (I-R) m Watermarked images I m
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11 The Proposed method − Lossless Recovery (1/2) To find S c which satisfy |D c |=|S|-|S c | |D c | = 3 001001011 010010010 001010011 111110 110R’100 001110 010010101 011010010 001011110 LSB of (I-R) m S c =001001011 S=000010101011 |S c |=9, |S|=12 O 001001011 D c = 010 010 000010101 011
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12 The Proposed method − Lossless Recovery (2/2) 149146148 144142144 146148 R 011 001 010 W 100101011001001010010100 100100001000111010010000 1001001010010100 111 111 111 D D c = 010 Extraction Decompression 21146148 1614144 1614620 R’ 000101011001001010010100 000100000000111010010000 000100001001001000010100
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13 Experimental results (1/2) Fig. 2. Visibly watermarked images with the MSB plane of R as R D (upper row) and the second MSB plane of R as R D (lower row), respectively.
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14 Experimental results (2/2) TABLE I: Performance evaluation. The unit of |D|; |D c |; and |S| is Bytes. N R D and N kb denote the bit plane level of R D and the key bit plane, respectively. The PSNR is calculated without R. (Unit: DB)
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15 Conclusions Design for binary watermarks The first work that implements a reversible visible watermarking system.
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16 Our proposed method- embedding (1/2) 11111 55555 222201113 33971804 44333 Host Image I Histogram peakzero 11111 66666 222201113 33971805 55333 a=3 b=6 01 10 Watermark W R I-R I’ D 10 01 92239 22552
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11111 66666 22922393 33225525 55333 17 Our proposed method- embedding (2/2) D= 1001 Watermarked images I m I’ 11111 66666 22922394 33225525 55433 a=3 b=6
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